Question

Find the area enclosed by the curves 3x² + 5y = 32 and y = | x − 2 |.

Answer 1

$3x^2+5y=32$ represents a downward opening parabola, symmetrical about negative x-axis

$$\begin{gathered} \mathrm{vertex}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{is}D\left(0,\frac{32}{5}\right).\mathrm{It}\mathrm{cuts}\mathrm{the}\mathrm{X}\mathrm{axis}\mathrm{at}B\left(\sqrt{\frac{32}{3}},0\right)\mathrm{and}B\text{'}\left(-\sqrt{\frac{32}{3}},0\right) \\ \mathrm{Also}, \\ y=\left|x-2\right| \\ \Rightarrow y=\left\{\begin{array}{l}x-2,x\ge 2\\ 2-x,x<2\end{array}\right. \\ y=x-2,x\ge 2\mathrm{represents}\mathrm{a}\mathrm{line},\mathrm{cutting}\mathrm{the}x\mathrm{axis}\mathrm{at}A\left(2,0\right)\mathrm{and}\mathrm{the}\mathrm{parabola}\mathrm{at}C\left(3,1\right) \\ \mathrm{And}y=2-x,x<2\mathrm{represents}\mathrm{a}\mathrm{line},\mathrm{cutting}\mathrm{the}\mathrm{parabola}\mathrm{at}\left(-2,4\right) \\ \mathrm{Shaded}\mathrm{area}\mathrm{A}EYCA \\ ={\int }_{-2}^2\left[\left(\frac{32-3x^2}{5}\right)-\left(2-x\right)\right]dx+{\int }_2^3\left[\left(\frac{32-3x^2}{5}\right)-\left(x-2\right)\right]dx \\ ={\int }_{-2}^3\left(\frac{32-3x^2}{5}\right)dx-\left[{\int }_{-2}^2\left(2-x\right)dx+{\int }_2^3\left(x-2\right)dx\right] \\ =\frac{1}{5}{\left[32x-x^3\right]}_{-2}^3-{\left[2x-\frac{x^2}{2}\right]}_{-2}^2-{\left[\frac{x^2}{2}-2x\right]}_2^3 \\ =\frac{1}{5}\left[32\times 3-27+64-8\right]-\left(4-2+4+2\right)-\left(\frac{9}{2}-6-2+4\right) \\ =\frac{1}{5}\left[96-35+64\right]-\left(8\right)-\left(\frac{9}{2}-4\right) \\ =\frac{125}{5}-4-\frac{9}{2} \\ =25-4-\frac{9}{2} \\ =21-\frac{9}{2} \\ =\frac{42-9}{2} \\ =\frac{33}{2}\mathrm{sq}\mathrm{units} \end{gathered}$$